{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "4769054f-d675-4678-852f-945df5f1fc7d",
   "metadata": {},
   "source": [
    "<div style=\"text-align: right\" align=\"right\"><i>Peter Norvig, December 2025</i></div>\n",
    "\n",
    "# Advent of Code 2025\n",
    "\n",
    "I  enjoy doing the [**Advent of Code**](https://adventofcode.com/) (AoC) programming puzzles, so here we go for 2025!  \n",
    "\n",
    "This year I will be doing something different: I will solve each problem myself here, and then [**in another notebook**](Advent-2025-AI.ipynb) I will ask an AI Large Language Model to solve the same problem. Check out the differences!\n",
    "\n",
    "# Day 0\n",
    "\n",
    "I'm glad that [@GaryGrady](https://mastodon.social/@garygrady) is providing cartoons:\n",
    "\n",
    "<a href=\"https://x.com/garyjgrady\"><img src=\"https://pbs.twimg.com/media/Gdp709FW8AAq2_m?format=jpg&name=medium\" width=500 alt=\"Gary Grady cartoon\"></a>\n",
    "\n",
    "I start by loading up my [**AdventUtils.ipynb**](AdventUtils.ipynb) notebook (same as last time except for the `current_year`). On each day I will first parse the input (with the help of my `parse` utility function), then solve Part 1 and Part 2 (recording the correct answer with my `answer` function)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "d5f1da68-5da6-434d-a068-1d93497a86b3",
   "metadata": {},
   "outputs": [],
   "source": [
    "%run AdventUtils.ipynb\n",
    "current_year = 2025"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "37bc12b8-d0dc-4873-984c-6f09ae647229",
   "metadata": {},
   "source": [
    "# [Day 1](https://adventofcode.com/2025/day/1): Secret Entrance\n",
    "\n",
    "On Day 1 we meet an elf and learn that our task is to finish decorating the North Pole by December 12th. There will be 24 challenges along the way. Today we need to unlock a safe. The safe has a dial with 100 numbers. Our input for today is a sequence of left and right rotations; for example \"R20\" means move the dial right by 20 numbers and \"L13\" means move it left by 13 numbers. I'll use my `parse` utility function to parse each line of the input as an integer, after replacing each 'L' with a minus sign and each 'R' with a plus sign:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "ed911a15-addc-4c04-8546-2c9f37aee341",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 4780 strs of size 2 to 4:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "L20\n",
      "L13\n",
      "L16\n",
      "L16\n",
      "L29\n",
      "L7\n",
      "L48\n",
      "L48\n",
      "...\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Parsed representation ➜ 4780 ints:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "-20\n",
      "-13\n",
      "-16\n",
      "-16\n",
      "-29\n",
      "-7\n",
      "-48\n",
      "-48\n",
      "...\n"
     ]
    }
   ],
   "source": [
    "def parse_rotation(line: str): return int(line.replace('L', '-').replace('R', '+'))\n",
    "\n",
    "rotations = parse(day=1, parser=parse_rotation) "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7c98c883-d1dc-4d4e-9590-d47b1de000a0",
   "metadata": {},
   "source": [
    "Note that my `parse` function prints the first few input lines, then prints the parsed representation of these lines. That helps me debug and helps you the reader understand what is going on.\n",
    "\n",
    "<img src=\"https://files.mastodon.social/media_attachments/files/115/646/343/679/448/846/original/428b312ca88f62c4.jpg\" width=500 alt=\"Gary Grady cartoon\">\n",
    "\n",
    "### Part 1: How many times is the dial left pointing at 0 after any rotation in the sequence?\n",
    "\n",
    "Initially the safe's arrow is pointing at 50, and then we apply the rotations in order. We are asked how many of the rotations leave the dial pointing at 0. The `itertools.accumulate` function yields running totals of its input sequence, so we just have to count (quantify) how many times the running total of the rotations is 0 mod 100. I'll write a function to do that and run it on the input:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "86079e27-2912-4431-8608-7a110a115789",
   "metadata": {},
   "outputs": [],
   "source": [
    "def count_zeros(numbers, dial=100) -> int:\n",
    "    \"\"\"How many zeros (modulo `dial`) in the running partial sums of the numbers?\"\"\"\n",
    "    return quantify(total % dial == 0 for total in accumulate(numbers, initial=50))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "eb90f15c-8f87-4d94-b49c-fcaa53aab8ae",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1182"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "count_zeros(rotations)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b5f68a70-8465-4249-954c-cba2b78751d8",
   "metadata": {},
   "source": [
    "I submitted \"1182\" to AoC and saw it was correct, so I record the `answer` like this:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "8387c4bd-a0b7-46d7-b726-d2d4b383b3cf",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  1.1:   .0006 seconds, answer 1182            correct"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(puzzle=1.1, solution=1182, code=lambda: \n",
    "       count_zeros(rotations))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "12d5d111-8f69-4944-855b-54cca8215d7b",
   "metadata": {},
   "source": [
    "### Part 2: How many times does the dial point to 0 at any time?\n",
    "\n",
    "For Part 2 we need to count both when a rotation ends up at 0 and when the arrow passes 0 at any time during a rotation. For example, if the arrow points to 95, then only a \"R5\" or a \"L95\" would register a 0 in Part 1, but now a rotation of \"R10\" would also count because it passes 0 (as would any rotation of \"R5\" or larger, or \"L95\" or larger). \n",
    "\n",
    "I'll start with a simple but slow approach: treat a rotation of, say, -20 as 20 rotations of -1, and then use the same `count_zeros` function from part 1. (Note that `sign(r)` returns +1 for any positive input, and -1 for any negative input.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "bfb5bd2d-d768-47b3-9897-d28f83f2202a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  1.2:   .1367 seconds, answer 6907            correct"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(1.2, 6907, lambda:\n",
    "       count_zeros(sign(r) for r in rotations for _ in range(abs(r))))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "bb0f6906-369e-4b3c-8840-d5648d713942",
   "metadata": {},
   "source": [
    "That's a long run time for a Day 1 problem, so here's a faster method: I break each rotation down into a number of full circles and some remainder, then add the full circles to the count of zeros, and add one more if the remainder is at least as much as the distance to zero (in the appropriate direction): "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "e0b676d8-3d4f-4b36-835d-b045ece53bd6",
   "metadata": {},
   "outputs": [],
   "source": [
    "def zero_clicks(rotations, position=50, dial=100) -> int:\n",
    "    \"\"\"How many times does any click cause the dial to point at 0?\n",
    "    Count 1 if the rotation crosses the distance to 0,\n",
    "    and for large rotations, count abs(r) // 100 more.\"\"\"\n",
    "    zeros = 0\n",
    "    for r in rotations:\n",
    "        full_circles, remainder = divmod(abs(r), dial)\n",
    "        distance_to_0 = (dial - position if (r > 0 or position == 0) else position)\n",
    "        zeros += full_circles + (1 if remainder >= distance_to_0 else 0)\n",
    "        position = (position + r) % dial\n",
    "    return zeros"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "8969191e-bd50-4187-8e4b-64524bc8427a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  1.2:   .0009 seconds, answer 6907            correct"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(1.2, 6907, lambda:\n",
    "       zero_clicks(rotations))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f47b1f7a-c9d9-4b21-a28a-f1e4e73d4c0d",
   "metadata": {},
   "source": [
    "That's much faster, but the code is trickier, and indeed I initially had a **bug** in the `distance_to_0` computation: when the current position is 0 the distance should be 100: it takes a full rotation to get back to 0. My code initially claimed the distance was 0; adding `or position == 0` fixed that."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dd6908ae-1906-4687-a50d-a50293a1dad5",
   "metadata": {},
   "source": [
    "# [Day 2](https://adventofcode.com/2025/day/2): Gift Shop\n",
    "\n",
    "Today we're in the North Pole gift shop, and are asked to help the elves identify invalid product IDs on the items there. We're giving a list of ranges of product IDs of items in stock. Each range is a pair of integers separated by a dash, and the ranges are separated by commas:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "63bb5099-68ab-41b3-8d0c-2f409433b3f2",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 1 str of size 511:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "990244-1009337,5518069-5608946,34273134-34397466,3636295061-3636388848,8613701-8663602,573252-68 ...\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Parsed representation ➜ 35 tuples of size 2:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "(990244, 1009337)\n",
      "(5518069, 5608946)\n",
      "(34273134, 34397466)\n",
      "(3636295061, 3636388848)\n",
      "(8613701, 8663602)\n",
      "(573252, 688417)\n",
      "(472288, 533253)\n",
      "(960590, 988421)\n",
      "...\n"
     ]
    }
   ],
   "source": [
    "id_ranges = parse(day=2, parser=positive_ints, sections=lambda text: text.split(','))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a5463e74-a3a1-4c79-9497-a1e307ef81e2",
   "metadata": {},
   "source": [
    "<img src=\"https://files.mastodon.social/media_attachments/files/115/652/152/368/251/243/original/56e4ed8e5f24db96.jpg\" width=500 alt=\"GaryJGrady cartoon\">\n",
    "\n",
    "### Part 1: What is the sum of the invalid IDs?\n",
    "\n",
    "An invalid ID is defined as one that consists of a digit sequence repeated twice. So 55, 6464 and 123123 are invalid. We're asked for the sum of the invalid IDs across all the ID ranges.\n",
    "\n",
    "We could look at every number in each range and check if the first half of the number (as a string) is the same as the second half. How many checks would that be?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "id": "d682f3f2-415e-4556-b6c8-43e331a38703",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1990936"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sum((hi - lo + 1) for lo, hi in id_ranges)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a9bf1aab-bd09-49a9-ab57-8b26b1f4a98d",
   "metadata": {},
   "source": [
    "Only 2 million! So it would indeed be feasible to check every one. But I have a suspicion that Part 2 would make it infeasible, so I'll invest in a more efficient approach. For each ID range, instead of enumerating every number in the range and checking each one for validity, I will instead enumerate over the *first half* of the possible digit strings, and automatically generate invalid IDs by appending a copy of the first half to itself. By *first half* I don't mean divide by 2; I mean the first half of the digit string: the first half of \"123456\" is \"123\".\n",
    "\n",
    "Suppose the range is 123456-223000.  I enumerate from 123 to 223, and for each number generate an invalid ID:\n",
    "[123123, 124124, 125125, ... 223223]. I then yield the IDs that are within the range (in this case all but the first and the last are in the range 123456-223000). Altogether I only have to consider 101 IDs rather than 100,001. (The algorithm scales with the square root of the size of the range, not with the size of the range itself.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "id": "1345f93d-84c5-43f8-b6c2-9fc7b8f5ed90",
   "metadata": {},
   "outputs": [],
   "source": [
    "def invalids_in_range(lo: int, hi: int) -> Iterable[int]:\n",
    "    \"\"\"Yield all the invalid IDs between lo and hi inclusive.\n",
    "    An ID is invalid if it consists of a digit sequence repeated twice.\"\"\"\n",
    "    first_half = str(lo)[:max(1, len(str(lo)) // 2)]\n",
    "    for i in count_from(int(first_half)):\n",
    "        id = int(str(i) * 2)\n",
    "        if lo <= id <= hi:\n",
    "            yield id\n",
    "        elif id > hi:\n",
    "            return\n",
    "\n",
    "def invalids(id_ranges) -> List[int]:\n",
    "    \"\"\"Invalid IDs, according to any one of the list of invalid ID ranges.\"\"\"\n",
    "    return append(invalids_in_range(lo, hi)\n",
    "                  for (lo, hi) in id_ranges)\n",
    "\n",
    "assert invalids([(11, 22)]) == [11, 22]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "id": "8b0d12b3-f184-4149-8b49-f9ff78663d46",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  2.1:   .0029 seconds, answer 23560874270     correct"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(2.1, 23560874270, lambda:\n",
    "       sum(invalids(id_ranges)))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "188dd774-0c2d-4304-8351-dfe208913a99",
   "metadata": {},
   "source": [
    "### Part 2: What is the sum of the invalid IDs, under the new rules?\n",
    "\n",
    "In Part 2 we discover that an ID should be considered invalid if it consists of two *or more* repeats of a sequence of digits. So 111 (1 repeated three times), 12121212 (12 repeated four times), and 222222 (2 repeated six times) are all invalid. I'll rewrite `invalids_in_range` to take an optional argument saying how many repeats we're looking for, and introduce  `all_invalids` to try all possible repeat lengths:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "id": "4fbf52e7-a06a-4000-8627-80d159788da1",
   "metadata": {},
   "outputs": [],
   "source": [
    "def invalids_in_range(lo: int, hi: int, repeat=2) -> Iterable[int]:\n",
    "    \"\"\"Yield all the invalid IDs between lo and hi inclusive\n",
    "    that are formed from exactly `repeat` repeated digit sequences.\"\"\"\n",
    "    first_half = int(str(lo)[:len(str(lo)) // repeat] or 1)\n",
    "    for i in count_from(int(first_half)):\n",
    "        id = int(str(i) * repeat)\n",
    "        if lo <= id <= hi:\n",
    "            yield id\n",
    "        elif id > hi:\n",
    "            return\n",
    "\n",
    "def all_invalids(id_ranges) -> Set[int]:\n",
    "    \"\"\"All Invalid IDs, according to the list of ranges, with any number of repeats.\"\"\"\n",
    "    return union(invalids_in_range(lo, hi, repeat)\n",
    "                 for (lo, hi) in id_ranges\n",
    "                 for repeat in range(2, len(str(hi)) + 1))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1e7d71d2-8c08-4673-8d12-661edd6ab6f5",
   "metadata": {},
   "source": [
    "Since this is a bit tricky, I include some unit test assertions:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "id": "5a29e5bc-3b07-4800-9a0f-4bd433e01c55",
   "metadata": {},
   "outputs": [],
   "source": [
    "assert list(invalids_in_range(2121212118, 2121212124, 5)) == [2121212121]\n",
    "assert all_invalids([(11, 22), (95, 115)]) == {11, 22, 99, 111}"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cb254f91-8c6d-445d-86be-97df9a93018c",
   "metadata": {},
   "source": [
    "Now I'll verify that the answer for Part 1 still works, and go ahead and compute the answer for Part 2:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "id": "8a7c6c25-4b5f-4178-8559-166ba1a9f924",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  2.1:   .0032 seconds, answer 23560874270     correct"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(2.1, 23560874270, lambda:\n",
    "       sum(invalids(id_ranges)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "id": "32fefd65-df2a-4ea3-9acd-7525ebd32380",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  2.2:   .0037 seconds, answer 44143124633     correct"
      ]
     },
     "execution_count": 16,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(2.2, 44143124633, lambda:\n",
    "       sum(all_invalids(id_ranges)))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "872cf212-bfbf-4edd-b898-5f76ad122a85",
   "metadata": {},
   "source": [
    "I initially had another **bug** here: I counted \"222222\" three times: once as 2 repeats of \"222\", once as 3 repeats of \"22\", and once as 6 repeats of \"2\". I changed the output of `all_invalids` to be a `set` to fix that."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "16e222aa-d424-44d8-bcb3-7aa11c31b540",
   "metadata": {},
   "source": [
    "# [Day 3](https://adventofcode.com/2025/day/3): Lobby\n",
    "\n",
    "Entering the lobby, we find that the elevators are offline. We might be able to fix the problem by turning on some batteries. There are multiple battery banks, each bank consisting of a sequence of batteries, each labeled with its *joltage*, a digit from 1 to 9. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "id": "04ef4f2f-2b07-43cd-87d4-a8f63f7fe840",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 200 strs of size 100:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "5353323523322232362334333433323333353233331313222372133133353643423323233323333534414523333432223242\n",
      "6344544745655555456556556566665564538465555575558846455665837545764555554465564547547565544657585435\n",
      "2246273372253242254243532252231242225522622633532222322234255122531222423531343223123232234213323424\n",
      "6545643634344444495734739454433454439454355654483544243344534445434437426443854344454534654439534424\n",
      "2356636643143433535443636338231745346538433576334436353176353333433532345344334224435234343644332536\n",
      "3221311221443323323322222214632342232233222322333436263122265162212321261323142262212332322125216222\n",
      "3336332333336335335324359336493238433441666379243536334165623214253384333323893933867663434332383763\n",
      "3235321252332431332223232436222532432226223222213233432853535322314122221322352235213323124321222233\n",
      "...\n"
     ]
    }
   ],
   "source": [
    "banks = parse(day=3)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "529e9177-7e47-46fa-bdc0-b979205d2e72",
   "metadata": {},
   "source": [
    "### Part 1: What is the maximum possible total output joltage?\n",
    "\n",
    "We can turn on exactly two batteries in each bank, resulting in a two digit number which is the *joltage* of the bank. For example, given the bank \"8647\" we could choose to turn on the \"8\" and \"7\" to produce a joltage of 87; that's the maximum.  The function `joltage` chooses the biggest first digit, and then the biggest second digit that follows the first digit. Note that the possible choices for the first digit exclude the last digit, because if we chose that, then there would be no choices left for the second digit. (I chose to do the string-to-int conversion in `total_joltage`; it would also be fine to have `joltage` return an int.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "id": "0168ada2-b53e-4215-9e6a-2468759f095c",
   "metadata": {},
   "outputs": [],
   "source": [
    "def joltage(bank: str) -> str:\n",
    "    \"\"\"The maximum possible joltage by turning on 2 batteries in the bank.\n",
    "    Pick the biggest first digit, then the biggest digit that follows.\"\"\"\n",
    "    first  = max(bank[:-1]) # The first digit can't be the last character\n",
    "    second = max(bank[bank.index(first) + 1:]) # The second digit must come after the first\n",
    "    return first + second\n",
    "\n",
    "def total_joltage(banks: List[str]) -> int:\n",
    "    \"\"\"The maximum possible joltage from all the banks.\"\"\"\n",
    "    return sum(int(joltage(bank)) for bank in banks)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "id": "3f6390f5-78b7-4472-acf9-8855d2c2453e",
   "metadata": {},
   "outputs": [],
   "source": [
    "assert joltage(\"8647\") == \"87\"\n",
    "assert joltage(\"1119\") == \"19\""
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "id": "9cc62ae9-b313-4b82-b8fb-bab9c0ef5cb6",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  3.1:   .0007 seconds, answer 17085           correct"
      ]
     },
     "execution_count": 20,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(3.1, 17085, lambda:\n",
    "       total_joltage(banks))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3f8a4645-5e36-4b3d-8ace-125024403b3b",
   "metadata": {},
   "source": [
    "### Part 2: What is the new maximum possible total output joltage?\n",
    "\n",
    "In Part 2 the elf hits the \"*joltage limit safety override*\" button, and we can now turn on 12 batteries per bank, resulting in a 12-digit joltage. What is the new maximum possible total joltage?\n",
    "\n",
    "I will make a change to the two functions, passing in the number of digits to be chosen, *n* (with default 2 for backwards compatibility). The function `joltage` recurses when there is more than one digit remaining, choosing the first digit from the bank up to the last *n* - 1 characters, then recursively finding the biggest joltage from the rest. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "id": "b9507c12-421b-41af-b926-a4d3d83f622d",
   "metadata": {},
   "outputs": [],
   "source": [
    "def joltage(bank: str, n=2) -> str:\n",
    "    \"\"\"The maximum possible joltage by turning on `n` batteries in the bank.\n",
    "    Pick the first digit, then the maximum joltage from the rest of the bank.\"\"\"\n",
    "    if n == 1:\n",
    "        return max(bank)\n",
    "    else:\n",
    "        first = max(bank[:-(n - 1)]) # The first digit can't be the last n-1 characters\n",
    "        rest  = bank[bank.index(first) + 1:]\n",
    "        return first + joltage(rest, n - 1)\n",
    "\n",
    "def total_joltage(banks: List[str], n=2) -> int:\n",
    "    \"\"\"The maximum possible joltage from all the banks,\n",
    "    when `n` batteries are turned on per bank.\"\"\"\n",
    "    return sum(int(joltage(bank, n)) for bank in banks)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "id": "4a58a657-8f32-4001-a4ff-f30dfbe7e5e7",
   "metadata": {},
   "outputs": [],
   "source": [
    "assert joltage(\"811111111111119\", 2)  == '89'\n",
    "assert joltage(\"818181911112111\", 5)  == '92111'\n",
    "assert joltage(\"818181911112111\", 12) == '888911112111'"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "28d2e2ce-98b5-4b3b-9f4d-26efcb7e1258",
   "metadata": {},
   "source": [
    "I'll make sure Part 1 still works, and then solve Part 2:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "id": "fe3cff78-81c0-4d4a-bb4d-f0f841067e0d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  3.1:   .0006 seconds, answer 17085           correct"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(3.1, 17085, lambda:\n",
    "       total_joltage(banks))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "id": "f971839e-81ea-49b4-a92f-a44884be645d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  3.2:   .0020 seconds, answer 169408143086082 correct"
      ]
     },
     "execution_count": 24,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(3.2, 169408143086082, lambda:\n",
    "       total_joltage(banks, 12))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "db13a440-9ad0-4344-b555-aec969869688",
   "metadata": {},
   "source": [
    "# [Day 4](https://adventofcode.com/2025/day/4): Printing Department\n",
    "\n",
    "The floor of the printing department is divided into squares, some of which contain a roll of paper. The day's input is a map of the floor, with `@` representing a roll of paper. I can handle that with the `Grid` class from my AdventUtils:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "id": "c9f227d5-2748-48e1-80c1-7385dad46323",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 140 strs of size 140:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      ".@@@@@...@.@@@@@@@@@@.@@@@@@@.@.@.@@@@@@@@@@@@@..@.@@@.@@@@@@..@.@..@.@@...@.@@@@..@@@@....@@@.@ ...\n",
      ".@@@@@.@....@.....@@@.@@.@@.@@@.@@@.@.@.@.@@@@@.@@.@@@@@.@@@@@@@@@@@..@@.@.@@.@@@.@@@@@@@@@@@..@ ...\n",
      "@.@@@@.@@@@.@@@@..@@.@@@@@@@@.@@@@.@@@@.@@..@.@...@.@.@.@.@@..@@@@@.@.@.@@@@.@@@@@@@@@.@@@@..@@. ...\n",
      ".@.....@.@@@..@.@@@.@..@@@@@..@@@.@@..@...@.@@@@.@@@.@.@@@@@@.@.@@@@@@@.@.@@@.@@@@@@...@@.@@..@. ...\n",
      "@@@@@.@@@.@@@@@@@..@@.@.@@@..@@..@@@.@@....@.@..@@@@@@@@.@.@@..@@...@@.@@@...@.@.@@@..@.@.@@@@@@ ...\n",
      "@.@@@@@@..@@@@...@..@@@@@@.@@@..@.....@@.@.@@...@@@.@@.@.@@@....@@.@.@.@@@@.@@@@@.@@@.@@...@@.@@ ...\n",
      ".@@@.@.@@@..@@.@.@@@@@.@.@..@@....@..@.@.@@@@.@..@@.@..@@@@@.@@@@@@@.@.@@@.@.@@@.@@@@.@@@@@@@@.@ ...\n",
      "@@@@@@@.@@...@@@....@.@@@@.@@@@@@@@@.@@@.@@.@@..@...@@@@@.@@@..@.@@@@@@@@@@.@@@.@..@@@.@@@@.@.@@ ...\n",
      "...\n"
     ]
    }
   ],
   "source": [
    "paper_grid = Grid(parse(day=4), directions=directions8)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4dd00e21-228c-41f6-a28c-e2213e60d4ce",
   "metadata": {},
   "source": [
    "<img src=\"https://files.mastodon.social/media_attachments/files/115/663/375/845/834/867/original/4d3180e12fa726fe.jpg\" width=500 alt=\"Gary Grady cartoon\">\n",
    "\n",
    "\n",
    "### Part 1: How many rolls of paper can be accessed by a forklift?\n",
    "\n",
    "A roll is **accessible** by forklift if there are fewer than four rolls of paper in the eight adjacent positions (that's why I specified `directions8` in defining `paper_grid`). Counting the number of accessible rolls is easy, but I decided to make `accessible rolls` return a list of positions rather than a count, in anticipation of Part 2."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "id": "6fe5bd44-8a28-4d7a-bc8b-edc4af8f23c3",
   "metadata": {},
   "outputs": [],
   "source": [
    "def accessible_rolls(grid: Grid) -> List[Point]:\n",
    "    \"\"\"The positions of all the accessible rolls of paper on the grid.\"\"\"\n",
    "    return [p for p in grid if is_accessible(p, grid)]\n",
    "\n",
    "def is_accessible(p: Point, grid: Grid) -> bool:\n",
    "    \"\"\"A roll of paper is accessible if there are fewer than \n",
    "    four rolls of paper in the eight adjacent positions.\"\"\"\n",
    "    return grid[p] == '@' and grid.neighbor_contents(p).count('@') < 4"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "714b9eed-bdee-4e3f-b0be-1a3f727dddfb",
   "metadata": {},
   "source": [
    "Here's the answer:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "id": "a5ef09cf-b204-41eb-80d8-de107d385dbb",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  4.1:   .0527 seconds, answer 1569            correct"
      ]
     },
     "execution_count": 27,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(4.1, 1569, lambda:\n",
    "       len(accessible_rolls(paper_grid)))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "550312ae-a70b-405c-90e6-5df9c9afc0e8",
   "metadata": {},
   "source": [
    "### Part 2: How many rolls of paper can be removed?\n",
    "\n",
    "If the elves can access a paper roll, they can remove it by forklift. That may in turn make other rolls accessible, and hence removable. How many rolls in total can be removed?\n",
    "\n",
    "It looks like I was right to make `accessible_rolls` return a list of points rather than a count! I can answer the question by repeatedly finding the accessible rolls, removing them (on a copy of the grid so I don't mess up the original grid), and repeating until there are no more accessible rolls."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "id": "0ed53853-268c-4c2f-a929-cb3e6005a348",
   "metadata": {},
   "outputs": [],
   "source": [
    "def removable_rolls(grid: Grid) -> Iterable[Point]:\n",
    "    \"\"\"The positions of paper rolls that can be removed, in any nuber of iterations.\"\"\"\n",
    "    grid2 = grid.copy() # To avoid mutating the input grid\n",
    "    points = accessible_rolls(grid2)\n",
    "    while points:\n",
    "        yield from points\n",
    "        grid2.update({p: '.' for p in points})\n",
    "        points = accessible_rolls(grid2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "id": "2fb17a51-05f7-42ec-8d6c-222121a026cf",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  4.2:  1.2384 seconds, answer 9280            correct"
      ]
     },
     "execution_count": 29,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(4.2, 9280, lambda:\n",
    "       quantify(removable_rolls(paper_grid)))     "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7143f73e-3b9b-49f3-bfa9-625899a56e37",
   "metadata": {},
   "source": [
    "That's the right answer, but the run time is slow. The main issue is that `accessible_rolls` has to look at the whole grid on every iteration. That's wasteful: If the previous iteration only removed one roll, all we really need to look at on the next iteration is the neighbors of the removed roll. So I'll keep a queue of possibly removable points and repeatedly pop a point off the queue, and if it is an accessible roll, remove it and put all its neighbors on the queue. When the queue is empty, no more rolls can be removed."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "id": "54f20b5e-6713-459c-8d40-e545ce6b8e42",
   "metadata": {},
   "outputs": [],
   "source": [
    "def removable_rolls(grid: Grid) -> Iterable[Point]:\n",
    "    \"\"\"The positions of paper rolls that can be removed, in any nuber of iterations.\"\"\"\n",
    "    grid2 = grid.copy() # To avoid mutating the original input grid\n",
    "    Q = list(grid)      # A queue of possibly removable positions in the grid\n",
    "    while Q:\n",
    "        p = Q.pop()\n",
    "        if is_accessible(p, grid2):\n",
    "            yield p\n",
    "            grid2[p] = '.'\n",
    "            Q.extend(grid2.neighbors(p))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "id": "bcba970b-09aa-479b-9c6d-4f6a7ac49fed",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  4.2:   .1424 seconds, answer 9280            correct"
      ]
     },
     "execution_count": 31,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(4.2, 9280, lambda:\n",
    "       quantify(removable_rolls(paper_grid)))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1f9a1e40-192a-4386-8cfb-bc5f69c88a9b",
   "metadata": {},
   "source": [
    "# [Day 5](https://adventofcode.com/2025/day/5): Cafeteria\n",
    "\n",
    "Today we're in the cafeteria, and the elves need to figure out which of their ingredients are fresh or spoiled. The input file has two parts (paragraphs), the first consisting of ranges of fresh ingredient IDs, like \"3-5\" and the second consisting of the available ingredient IDs, like \"8\". I can parse the data like this:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "id": "4b508db5-aeae-410e-a062-1b2d4ce41253",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 1184 strs of size 0 to 31:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "292632986393425-296797126337251\n",
      "428261559408337-431275643240865\n",
      "197704206528056-198822557342819\n",
      "36791726875734-37049023408764\n",
      "134880223152389-139959748438608\n",
      "31870818340663-32138457068292\n",
      "443613579514078-447616030778273\n",
      "284142407577672-288117756872436\n",
      "...\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Parsed representation ➜ 2 tuples of size 183 to 1000:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "((292632986393425, 296797126337251), (428261559408337, 431275643240865), (197704206528056, 19882 ...\n",
      "(92087202605588, 314304587960778, 19184152482180, 331502336245673, 104093299503920, 224082459481 ...\n"
     ]
    }
   ],
   "source": [
    "def parse_ingredients(text: str) -> tuple:\n",
    "    \"\"\"Parse either ranges like \"3-5\", or integers like \"8\".\"\"\"\n",
    "    return mapt(positive_ints, lines(text)) if '-' in text else ints(text)\n",
    "                                                   \n",
    "fresh_ranges, available_ingredient_ids = parse(day=5, parser=parse_ingredients, sections=paragraphs)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ebb61d5f-4683-4ff6-986c-69154d512e1b",
   "metadata": {},
   "source": [
    "### Part 1: How many of the available ingredient IDs are fresh?\n",
    "\n",
    "Keep it simple: For each ingredient ID, check to see if it is contained in any of the ranges (then we'll see if this is fast enough):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "id": "ae5b33e4-92f5-4fe2-bd5a-6b22095724fa",
   "metadata": {},
   "outputs": [],
   "source": [
    "def count_fresh_ingredients(fresh_ranges, available_ingredient_ids) -> int:\n",
    "    \"\"\"How many of the available ingredient IDs are in one of the fresh ranges?\"\"\"\n",
    "    return quantify(any(lo <= id <= hi for (lo, hi) in fresh_ranges)\n",
    "                    for id in available_ingredient_ids)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "id": "112a84f7-9bb8-45f0-9d7e-f600f37f0fdf",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  5.1:   .0079 seconds, answer 635             correct"
      ]
     },
     "execution_count": 34,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(5.1, 635, lambda:\n",
    "       count_fresh_ingredients(fresh_ranges, available_ingredient_ids))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "01c2c322-e873-412f-9c47-59ffdf9adf79",
   "metadata": {},
   "source": [
    "Fast enough! Let's move on.\n",
    "\n",
    "### Part 2: How many ingredient IDs are fresh?\n",
    "\n",
    "In Part 2 we are asked how many of the possible ingredient IDs are fresh, regardless of whether the ID is available or not. I could convert a range like \"3-5\" into the set {3, 4, 5} and then union the sets together. How many elements would be in that set? Here's an estimate: "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "id": "47a660d2-746c-433f-b4ae-9d01f70bc504",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "476036797138761"
      ]
     },
     "execution_count": 35,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sum(hi - lo for (lo, hi) in fresh_ranges)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1360869f-3ccb-4c71-8d00-35de1215389f",
   "metadata": {},
   "source": [
    "OK, switch to Plan B. Instead of explicitly building this set I can instead sort the fresh ID ranges (by their low number) and go through them, keeping track of the lowest ID that has not yet been explored for freshness, and a count of the fresh IDs found so far. For each range, the number of new fresh IDs is the length of the range that starts with either the start of the range or the first unexplored ID, and ends at the end of the range."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "id": "f52e8ecd-325e-4ed4-8928-de9365b5b7d4",
   "metadata": {},
   "outputs": [],
   "source": [
    "def count_fresh_ids(fresh_ranges) -> int:\n",
    "    \"\"\"How many IDs are contained in the union of the ranges?\"\"\"\n",
    "    fresh_count = 0 # The number of fresh IDs found so far\n",
    "    unexplored  = 0 # The highest ID number that we haven't considered yet\n",
    "    for (lo, hi) in sorted(fresh_ranges):\n",
    "        start = max(unexplored, lo)\n",
    "        fresh_count += len(range(start, hi + 1))\n",
    "        unexplored = max(unexplored, hi + 1)\n",
    "    return fresh_count"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "id": "fab6f1ed-543f-4f42-b1d5-2551742e3a4f",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  5.2:   .0001 seconds, answer 369761800782619 correct"
      ]
     },
     "execution_count": 37,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(5.2, 369761800782619, lambda:\n",
    "       count_fresh_ids(fresh_ranges))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6b83e353-3b77-4645-85b9-2f6ee6ff9e1d",
   "metadata": {},
   "source": [
    "# [Day 6](https://adventofcode.com/2025/day/5): Trash Compactor\n",
    "\n",
    "Trash Compactor? [I've got a bad feeling  about this!](https://youtu.be/CZgeYSqUeTA?si=5UPS_HiCOmTKrEWX&t=32) We've fallen into a garbage smasher and have been asked to help some of the resident cephalopods children with their math homework. We can parse the homework worksheet, but we were told that the exact alignment of columns matters, so I'll keep each line as a string rather than converting it to a list of ints.\n",
    "\n",
    "<img src=\"https://files.mastodon.social/media_attachments/files/115/674/362/938/836/432/original/3b83547aa749a914.jpg\" width=500 alt=\"Gary Grady cartoon\">"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "id": "bca852a9-e4c5-4706-abfc-afc6d4a4eeb5",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 5 strs of size 3732:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "   4  82 68 85 74 6  56 14 2  8669 66 13 927  3 235 44   52 16  37  61 82 1774  96 15 775 789 91 ...\n",
      "   6  87 39 72 56 12 69 79 58 4378 86 49 146  5 412 85 7751 577 69 813 55 9942 753 49 734 587 15 ...\n",
      " 827 446 82 72 76 21 31 32 96 3257 21 21 171  2 31  17 9178 977 11 469 58  712 162 4    1 132 91 ...\n",
      "9472 154 36 76  5 89 37  5 28    6 95 49  82 66 7   44 8183 524 38 819 11    7 825 9    8 397 48 ...\n",
      "+    +   +  *  *  *  +  *  *  +    *  +  *   *  +   +  +    +   *  +   *  +    *   +  *   *   *  ...\n"
     ]
    }
   ],
   "source": [
    "worksheet = parse(6, str)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ae8a7f1d-07ec-46e4-880f-abbd82d46a67",
   "metadata": {},
   "source": [
    "### Part 1: What is the grand total of the answers to the individual problems?\n",
    "\n",
    "We humans are used to arithmetic problems written in left-to-right order, but cephalopods use top-to-bottom order, with a postfix operator in the last row. We need to compute the sum or product for each column and add all those results together. So the problem in the first column above would be \"`4 + 6 + 827 + 9472`\". Here's the code:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "id": "92f320f4-c7a3-4dfd-ae61-6b1b180bdc93",
   "metadata": {},
   "outputs": [],
   "source": [
    "def grand_total(worksheet):\n",
    "    \"\"\"The sum of the individual sum/product problems, where each column is a problem.\"\"\"\n",
    "    columns = T(map(str.split, worksheet)) # Columns are the transpose of the rows; `T` is transpose\n",
    "    operations = {'*': prod, '+': sum}\n",
    "    return sum(operations[op](mapt(int, numbers)) \n",
    "               for (*numbers, op) in columns)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "id": "92a47f3a-127f-41e8-bee1-6276885bd36b",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  6.1:   .0015 seconds, answer 5877594983578   correct"
      ]
     },
     "execution_count": 40,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(6.1, 5877594983578, lambda:\n",
    "       grand_total(worksheet))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "226772ae-0d6e-43b8-9159-171ec1b36a5d",
   "metadata": {},
   "source": [
    "### Part 2: What is the grand total of the answers to the individual problems with the new rules?\n",
    "\n",
    "We learn that we did all the problems wrong. Cephalopodish number notation is different; numbers are read vertically rather than horizontally and the exact column alignment of each digit matters. Given the worksheet:\n",
    "\n",
    "       4  82 68 85 74 6  56 14 2  8669 66 13 927  3 235 44 \n",
    "       6  87 39 72 56 12 69 79 58 4378 86 49 146  5 412 85 \n",
    "     827 446 82 72 76 21 31 32 96 3257 21 21 171  2 31  17 \n",
    "    9472 154 36 76  5 89 37  5 28    6 95 49  82 66 7   44 \n",
    "    +    +   +  *  *  *  +  *  *  +    *  +  *   *  +   +  \n",
    "\n",
    "The problem in the leftmost column is not \"`4 + 6 + 827 + 9472`\"; rather it is \"`9 + 84 + 27 + 4672`\".\n",
    "\n",
    "That means I can't just split each line into numbers, I'll have to be careful to maintain the blank spaces to the right and left of the digits, and I have to know in what position each column starts and ends. That part was tricky, so here's an explanation:\n",
    "- I note from the worksheet above that each column starts at a position above  a `+` or `*` sign.\n",
    "- Each column ends one character befpre the next `+` or `*` sign (one character is a blank space to separate columns).\n",
    "- For the last column there is no terminator, so I'll add the string `' *'` to the operator line before computing the start positions.\n",
    "\n",
    "In `grand_total2` I first break each line into columns, take the transpose of that (to give a list of problems), do the math on each problem, and return the sum of the results. Within `cephalopodish_math` I call `vertically`, whioch again does a transpose and then puts the digits together into an integer."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "id": "5866583b-245d-4a43-a037-229bf35f1be2",
   "metadata": {},
   "outputs": [],
   "source": [
    "def grand_total2(worksheet: List[str]):\n",
    "    \"\"\"Solve the problem in each column with cephalopodish math and return the sum.\"\"\"\n",
    "    problems = T(break_into_columns(worksheet))\n",
    "    return sum(map(cephalopodish_math, problems))\n",
    "\n",
    "def break_into_columns(worksheet: List[str]) -> List[List[str]]:\n",
    "    \"\"\"Break a whole worksheet into columns by finding the `blanks` (the columns with\n",
    "    only blank spaces) and then separating each line at these posisitons.\"\"\"\n",
    "    *number_lines, operator_line = worksheet\n",
    "    operator_line = operator_line.ljust(len(worksheet[0])) + ' *'\n",
    "    column_starts = [i for i, ch in enumerate(operator_line) if ch != ' ']\n",
    "    return [break_line_into_columns(line, column_starts) for line in worksheet]\n",
    "\n",
    "def break_line_into_columns(line: str, column_starts: List[int]) -> List[str]:\n",
    "    \"\"\"Break one line into columns, as specified by the list of `blanks` positions.\"\"\"\n",
    "    return [line[column_starts[i]:column_starts[i + 1]-1] \n",
    "            for i in range(len(column_starts) - 1)]\n",
    "\n",
    "def cephalopodish_math(problem) -> int:\n",
    "    \"\"\"Return the sum or product of the vertically-arranged numbers.\"\"\"\n",
    "    *numbers, op = problem\n",
    "    sum_or_prod = (sum if op.strip() == '+' else prod)\n",
    "    return sum_or_prod(vertically(numbers))\n",
    "\n",
    "def vertically(numbers: List[str]) -> List[int]:\n",
    "    \"\"\"Return a list of integers found by read numbers vertically by column.\"\"\"\n",
    "    return [int(cat(digits)) for digits in T(numbers)]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "id": "39a8fa78-946f-45aa-8d36-76883f4aeaff",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  6.2:   .0039 seconds, answer 11159825706149  correct"
      ]
     },
     "execution_count": 42,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(6.2, 11159825706149, lambda:\n",
    "       grand_total2(worksheet))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ca461ef3-50c3-4189-b724-5f2b3898f27d",
   "metadata": {},
   "source": [
    "I initially had an `IndexError` **bug** because the operator line is shorter then the numbers lines. Then I had an off-by-one **bug** that messed up the problem in the last column. To debug my errors I worked on the smaller example worksheet, doing things like this:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "id": "a13dbe89-3178-433f-9a50-28ed9a2f8358",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3263827"
      ]
     },
     "execution_count": 43,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "example = \"\"\"\\\n",
    "123 328  51 64 \n",
    " 45 64  387 23 \n",
    "  6 98  215 314\n",
    "*   +   *   +\"\"\".splitlines()\n",
    "\n",
    "grand_total2(example)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "id": "3aa0b661-367a-4316-905a-60f1d9e6df01",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[['123', '328', ' 51', '64 '],\n",
       " [' 45', '64 ', '387', '23 '],\n",
       " ['  6', '98 ', '215', '314'],\n",
       " ['*  ', '+  ', '*  ', '+']]"
      ]
     },
     "execution_count": 44,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "break_into_columns(example)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "id": "13392d9c-fef8-4946-b945-48ad6bc7eca9",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[('123', ' 45', '  6', '*  '),\n",
       " ('328', '64 ', '98 ', '+  '),\n",
       " (' 51', '387', '215', '*  '),\n",
       " ('64 ', '23 ', '314', '+')]"
      ]
     },
     "execution_count": 45,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "T(break_into_columns(example))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "id": "e29df77c-7385-4ba9-a27b-4074a78a0d74",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[1, 24, 356]"
      ]
     },
     "execution_count": 46,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "vertically(('123', \n",
    "            ' 45', \n",
    "            '  6'))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "id": "9a25ff50-f465-4de3-95ca-71b7a3d9569b",
   "metadata": {},
   "outputs": [],
   "source": [
    "assert cephalopodish_math(('123', ' 45', '  6', '*  ')) == 1 * 24 * 356 == 8544"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "953679f2-8b55-4bcb-b46f-db0d7c435ea3",
   "metadata": {},
   "source": [
    "# [Day 7](https://adventofcode.com/2025/day/7): Laboratories\n",
    "\n",
    "In the teleporter lab is a tachyon manifold in need of repair. We find a diagram of the manifold, with `S` marking the start of the tachyon beams and `^` marking is a splitter. \n",
    "\n",
    "I could parse this as a `Grid`, but I think I will just keep it as a list of strings. The main idea of a `Grid` is dealing with the 4 or 8 neighbors; a concept that this problem does not use."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "id": "e85af831-b6cf-4073-a131-ce68debaea75",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 142 strs of size 141:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "......................................................................S......................... ...\n",
      "................................................................................................ ...\n",
      "......................................................................^......................... ...\n",
      "................................................................................................ ...\n",
      ".....................................................................^.^........................ ...\n",
      "................................................................................................ ...\n",
      "....................................................................^.^.^....................... ...\n",
      "................................................................................................ ...\n",
      "...\n"
     ]
    }
   ],
   "source": [
    "manifold = parse(7)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "59927e26-3252-4c26-8aef-014afdb6deaf",
   "metadata": {},
   "source": [
    "### Part 1: How many times will the beam be split?\n",
    "\n",
    "Tachyon beams move downwards unless they enounter a `^` splitter, in which case they split into two beams, one immediately to the left of the splitter and one to the right. We're asked how many splits occur. If two beams end up in the same place they count as one beam, so if that beam is split again, that's just one more split, not two more.\n",
    "\n",
    "That suggests I should keep a `set` of current beam positions, and update the set every time we go down one line. (It is a set rather than a list so that duplicate beam positions count as one, not as two. (Also, if I kept a list of positions then given the size of the input there would be quadrillions of beams by the end, so it would take a very long time to get the wrong answer.)) I use the trick of iterating with `for b in list(beams)` rather than `for b in beams` so that I can mutate the set `beams` during the iteration. I also need to keep track of the split count, and return that at the end:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "id": "d4d3bdbb-d8b2-4e22-adb4-c99a212a2ca7",
   "metadata": {},
   "outputs": [],
   "source": [
    "def count_splits(manifold: List[str]) -> int:\n",
    "    \"\"\"How many beam split events occur as the beams make their way through the manifold?\"\"\"\n",
    "    start = manifold[0].index('S')\n",
    "    beams = {start}\n",
    "    split_count = 0\n",
    "    for line in manifold:\n",
    "        for b in list(beams):\n",
    "            if line[b] == '^': # Replace the beam position with one on each side and count a split\n",
    "                split_count += 1\n",
    "                beams.remove(b)\n",
    "                beams.update({b - 1, b + 1})\n",
    "    return split_count"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "id": "0e14ad5e-6871-4a9c-bd94-5a863c2341e0",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  7.1:   .0008 seconds, answer 1681            correct"
      ]
     },
     "execution_count": 50,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(7.1, 1681, lambda:\n",
    "       count_splits(manifold))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7e8174b9-2b66-41ce-9677-807a90b88ec5",
   "metadata": {},
   "source": [
    "### Part 2: In total, how many different timelines would a single tachyon particle end up on?\n",
    "\n",
    "Now we're told this is a *quantum* tachyon manifold and we need to know how many different *timelines* a single tachyon appears in, or in other words, how many different paths can the tachyon beams take to get to the last line.\n",
    "\n",
    "We can't just count the number of beams in the last line, because if a beam in position *b* takes two different paths to get there, that counts as two, not one. Instead, I'll replace the `set` from Part 1 with a `Counter` of `{beam_position: number_of_paths_to_get_here}`. If a beam at position *b* with *n* paths to get there encounters a splitter, then in the next step both positions *b* - 1 and *b* + 1 are incremented by *n*. In the end we return the sum of all the counts of paths."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "id": "c231059d-edad-4f1c-b584-e811fa8fad46",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  7.2:   .0014 seconds, answer 422102272495018 correct"
      ]
     },
     "execution_count": 51,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def count_timelines(manifold: List[str]) -> int:\n",
    "    \"\"\"How many possible paths are there to the final line of the manifold?\"\"\"\n",
    "    start = manifold[0].index('S')\n",
    "    beams = Counter({start: 1})\n",
    "    for line in manifold:\n",
    "        for b, n in list(beams.items()):\n",
    "            if line[b] == '^':\n",
    "                beams[b]     -= n\n",
    "                beams[b - 1] += n\n",
    "                beams[b + 1] += n\n",
    "    return sum(beams.values())\n",
    "\n",
    "answer(7.2, 422102272495018, lambda:\n",
    "       count_timelines(manifold))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ed2369ae-cf37-4b12-9933-4bfe95fec3ac",
   "metadata": {},
   "source": [
    "<img src=\"https://files.mastodon.social/media_attachments/files/115/680/410/219/564/454/original/13e14a398cee7827.jpg\" width=500 alt=\"Gary Grady cartoon\">"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "135b532d-d9ac-4400-9c6d-3b316d4596e1",
   "metadata": {},
   "source": [
    "# [Day 8](https://adventofcode.com/2025/day/8): Playground\n",
    "\n",
    "In a giant playground, some elves are setting up an ambitious Christmas light decoration project using small electrical junction boxes. They have a list of the junction box coordinates in 3D space:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "id": "a5cc43b2-3c5d-4d28-8c94-112bbb5739e7",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Puzzle input ➜ 1000 strs of size 13 to 17:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "83023,97914,59845\n",
      "95221,6604,47587\n",
      "15200,13848,533\n",
      "91106,47235,95124\n",
      "91396,58746,56994\n",
      "72070,33359,10785\n",
      "13733,43410,23707\n",
      "84853,78378,68689\n",
      "...\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "Parsed representation ➜ 1000 tuples of size 3:\n",
      "────────────────────────────────────────────────────────────────────────────────────────────────────\n",
      "(83023, 97914, 59845)\n",
      "(95221, 6604, 47587)\n",
      "(15200, 13848, 533)\n",
      "(91106, 47235, 95124)\n",
      "(91396, 58746, 56994)\n",
      "(72070, 33359, 10785)\n",
      "(13733, 43410, 23707)\n",
      "(84853, 78378, 68689)\n",
      "...\n"
     ]
    }
   ],
   "source": [
    "boxes = parse(8, ints)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ee9f47ae-09f5-479e-9e34-34246d723952",
   "metadata": {},
   "source": [
    "### Part 1: What is the product of the sizes of the three largest circuits?\n",
    "\n",
    "The goal is to start connecting junction boxes, starting with the two boxes that are closest to each other in 3D space, then the next two closest, and so on. (I assume that a box can connect to any number of other boxes.) After connecting the 1000 pairs that are closest together,  what do you get if you multiply together the sizes of the three largest circuits?\n",
    "\n",
    "I recognize this as a [**greedy algorithm**](https://en.wikipedia.org/wiki/Greedy_algorithm), consuming shortest links first, and I've [done that before](TSP.ipynb). I also recognize this as a [**Union-Find**](https://en.wikipedia.org/wiki/Disjoint-set_data_structure) problem, and I know there are efficient data structures for that problem we don't make heavy use of the union-find functionality, so keeping it simple seems to be the best result. (After finishing my code I feel vindicated:  I see in my [other notebook](Advent2025-AI.ipynb) that Claude Opus 4.5 also recognized the Union-Find problem, implemented a data structure for it, and ended up with code that ran slower than my simpler approach.)\n",
    "\n",
    "The function `greedy_connect` will keep a dict that maps each box to the circuit it is part of (initially just itself), and update that dict when two circuits are connected together. Then I'll go through the 1000 `closest_pairs` of boxes, updating the dict for each one, and return the dict at the end. Then the function `largest` will find the largest 3 circuits, and `prod` will multiply the sizes together."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "id": "fb7744c9-105b-439b-aa57-caf93b8117b8",
   "metadata": {},
   "outputs": [],
   "source": [
    "def greedy_connect(boxes, n=1000) -> Dict[Point, Tuple[Point, ...]]:\n",
    "    \"\"\"Go through the `n` closest pairs of boxes, shortest first. \n",
    "    If two boxes can be connected to form a new circuit, do it.\"\"\"\n",
    "    circuits = {B: (B,) for B in boxes} # A dict of {box: circuit}\n",
    "    for (A, B) in closest_pairs(boxes, n):\n",
    "        if circuits[A] != circuits[B]:\n",
    "            new_circuit = circuits[A] + circuits[B]\n",
    "            for C in new_circuit:\n",
    "                circuits[C] = new_circuit\n",
    "    return circuits\n",
    "\n",
    "def closest_pairs(points, n=1000) -> List[Tuple[Point, Point]]:\n",
    "    \"Return the `n` closest pairs of points, sorted shortest first.\"\n",
    "    pairs = combinations(points, 2)\n",
    "    return sorted(pairs, key=lambda link: distance_squared(*link))[:n]\n",
    "\n",
    "def largest(n, circuits) -> List[int]: \n",
    "    \"\"\"The lengths of the `n` largest circuits (ones with the most boxes).\"\"\"\n",
    "    return sorted(map(len, set(circuits.values())))[-n:]\n",
    "\n",
    "def distance_squared(P: Point, Q: Point) -> float:\n",
    "    \"\"\"The square of the distance between 3D points P and Q.\"\"\"\n",
    "    # Since we only care about sorting by distance, the distance squared is cheaper to compute\n",
    "    return (P[0] - Q[0]) ** 2 + (P[1] - Q[1]) ** 2 + (P[2] - Q[2]) ** 2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "id": "337ab7d6-d142-4c56-a0a3-2b6cd06cd895",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  8.1:   .6207 seconds, answer 24360           correct"
      ]
     },
     "execution_count": 54,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(8.1, 24360, lambda:\n",
    "       prod(largest(3, greedy_connect(boxes, 1000))))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3291fa28-a8fa-4ef8-90b3-2b4b6fc9625e",
   "metadata": {},
   "source": [
    "### Part 2: What do you get if you multiply together the X coordinates of the last two junction boxes you need to connect?\n",
    "\n",
    "Now we are asked for the product of the X coordinates of the final pair of boxes to be connected. Unfortunately I'll have to copy and paste most of `greedy_connect` to form `last_connected`, which checks to see if all the boxes are connected into one circuit, and if so returns the two boxes that made the final connection. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "id": "1096ded6-749e-4787-9b90-2de917b147c4",
   "metadata": {},
   "outputs": [],
   "source": [
    "def last_connected(boxes) -> dict:\n",
    "    \"\"\"Go through tall the pairs of boxes, in closest first order. \n",
    "    Return the last two boxes that finally make it all one big circuit.\"\"\"\n",
    "    circuits = {B: (B,) for B in boxes} # A dict of {box: circuit}\n",
    "    for (A, B) in closest_pairs(boxes, -1):\n",
    "        if circuits[A] != circuits[B]:\n",
    "            new_circuit = circuits[A] + circuits[B]\n",
    "            if len(new_circuit) == len(boxes):\n",
    "                return (A, B)\n",
    "            for C in new_circuit:\n",
    "                circuits[C] = new_circuit"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "id": "58f06244-ca67-47c9-8ca5-3346a249e1fc",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "Puzzle  8.2:   .6530 seconds, answer 2185817796      correct"
      ]
     },
     "execution_count": 56,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "answer(8.2, 2185817796, lambda:\n",
    "       prod(Xs(last_connected(boxes))))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "fa988909-1a8b-4e8c-aca6-c53af99bc0b6",
   "metadata": {},
   "source": [
    "Today's puzzles had the slowest run times yet. I could perhaps make them faster by mutating sets rather than forming a new tuple for each new circuit, but I think gains from that would be small, and since the run time is still under a second, I'll leave the code as is."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7f31ae9b-6606-40b0-9bb1-ed9b3fe3cbf0",
   "metadata": {},
   "source": [
    "# Summary\n",
    "\n",
    "Here are the run times and the correct answers for each of the puzzles:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "id": "ba36579c-d0b4-4fd3-939c-0026ecddd7e9",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Puzzle  1.1:   .0006 seconds, answer 1182            correct\n",
      "Puzzle  1.2:   .0009 seconds, answer 6907            correct\n",
      "Puzzle  2.1:   .0032 seconds, answer 23560874270     correct\n",
      "Puzzle  2.2:   .0037 seconds, answer 44143124633     correct\n",
      "Puzzle  3.1:   .0006 seconds, answer 17085           correct\n",
      "Puzzle  3.2:   .0020 seconds, answer 169408143086082 correct\n",
      "Puzzle  4.1:   .0527 seconds, answer 1569            correct\n",
      "Puzzle  4.2:   .1424 seconds, answer 9280            correct\n",
      "Puzzle  5.1:   .0079 seconds, answer 635             correct\n",
      "Puzzle  5.2:   .0001 seconds, answer 369761800782619 correct\n",
      "Puzzle  6.1:   .0015 seconds, answer 5877594983578   correct\n",
      "Puzzle  6.2:   .0039 seconds, answer 11159825706149  correct\n",
      "Puzzle  7.1:   .0008 seconds, answer 1681            correct\n",
      "Puzzle  7.2:   .0014 seconds, answer 422102272495018 correct\n",
      "Puzzle  8.1:   .6207 seconds, answer 24360           correct\n",
      "Puzzle  8.2:   .6530 seconds, answer 2185817796      correct\n",
      "\n",
      "Time in seconds: sum = 1.495, mean =  .093, median =  .003, max =  .653\n"
     ]
    }
   ],
   "source": [
    "summary(answers)"
   ]
  }
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